A topological space is called quasi-compact if every open cover of it has a finite subcover. Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. The usual proof that $Z$ is quasi-compact uses a maximal filter, hence Axiom of Choice. Can we prove it without using Axiom of Choice?

Edit(Apr. 14, 2014) If I am not mistaken, I have come up with a proof without using Axiom of Choice. I would like to post it as an answer.

Edit(Apr. 15, 2014) I was mistaken. As Andres Caicedo pointed out, I used AC without noticing it in my "proof".

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    $\begingroup$ @Arthur: Because it will incite another long circle of "why was this question closed", "why was this meta thread closed", "is it wrong to ask questions?", "why is there a group of users that votes against everything I do?" and so on and so forth. $\endgroup$
    – Asaf Karagila
    Apr 13, 2014 at 8:43
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    $\begingroup$ @AsafKaragila Stop it. You are abusing the thread by an off topic matter. $\endgroup$ Apr 13, 2014 at 8:48
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    $\begingroup$ @Makoto: Stop it. You are abusing the thread by an off topic matter. I see that I have to start quoting your comments too, in case that you delete them. For example, now it seems that my last comment is "out of the blue" and not at all in reply of you being all naive and innocent about your behavior. $\endgroup$
    – Asaf Karagila
    Apr 13, 2014 at 8:52
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    $\begingroup$ Please improve the question - as it stands, it shows no effort. For example, you could examine the usual proof that the product of two quasicompact spaces is compact, and point out where the axiom of choice is actually used. @OP $\endgroup$ Apr 13, 2014 at 12:23
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    $\begingroup$ @Makoto Kato: [[the issue is that you didn't do enough before asking the question.] How do you know that?] You asked already - see math.stackexchange.com/questions/753072/… and math.stackexchange.com/questions/753072/… . You appear to ignore responses such as that, while reading others in a way that makes them more favorable to you than they actually are (e.g. the fact that it's sometimes acceptable to answer one's own question). $\endgroup$ Apr 14, 2014 at 15:11

2 Answers 2


Here is a proof which is choice free. I should also mention that the proof using ultrafilters appears in Herrlich's The Axiom of Choice, where later he says that it can be modified to work without the axiom of choice using some ideas from other proofs. I did not check that claim in details, though.

First, let us prove the following lemma:

Lemma. Let $X$ be a topological space and $\cal B$ a basis for the topology. $X$ is quasi-compact if and only if every cover with elements of $\cal B$ has a finite subcover.

Proof. One direction is trivial, if $X$ is quasi-compact, certainly every cover with elements of $\cal B$ has a finite subcover. In the other direction, suppose that $\mathcal U=\{U_i\mid i\in I\}$ is an open cover, consider $\cal U'$ to be the refined cover, $\{V\in\mathcal B\mid\exists i\in I: V\subseteq U_i\}$. Then $\cal U'$ is an open cover as well, since every point in $U_i$ lies within some element of $\cal B$. Let $V_1,\ldots,V_n\in\cal U'$ be a finite subcover, then we can choose $U_1,\ldots,U_n\in\cal U$ such that $V_i\subseteq U_i$ for all $i\leq n$, and this is a finite subcover as wanted. $\ \square$

Now we can prove our theorem.

Let $X,Y$ be two quasi-compact spaces, and let $\cal U$ be an open covering of $X\times Y$ using rectangles (that is, sets of the form $U\times V$ where $U$ is open in $X$ and $V$ open in $Y$). We say that $A\subseteq X$ is adequate [for $\cal U$] if $A\times Y$ has a finite subcover in $\cal U$. Our goal is to show that $X$ is adequate, then $X\times Y$ can be covered by a finite subcover of $\cal U$.

First we show that if $x\in X$, then there is some $U\subseteq X$ which is open and $x\in U$ such that $U$ is adequate. Let $U_1\times V_1,\ldots,U_n\times V_n$ be a finite subcover such that $\{x\}\times Y\subseteq U_1\times V_1\cup\ldots\cup U_n\times V_n$ (such finite cover exists since $\{x\}\times Y$ is quasi-compact). Now consider $U=\bigcap_{i=1}^n U_i$, then $U$ is open as a finite intersection of open sets, and non-empty since $\{x\}\in U$, as wanted. $U$ is adequate since given any $(u,y)\in U\times Y$ we have that for some $i\leq n$ it is true that $(x,y)\in U_i\times V_i$, so $(u,y)\in U_i\times V_i$ as well (since $u\in U_i$). Therefore $U_1\times V_1,\ldots,U_n\times V_n$ is a cover of $U\times Y$.

Next we note that the finite union of adequate sets is adequate (as it is covered by the [finite] union of the [finite] subcovers of the adequate sets).

Now $\{U\subseteq X\mid U\text{ is open and adequate}\}$ is an open cover of $X$, by the above fact that every $x\in X$ has an adequate neighborhood, and by quasi-compactness of $X$ it has a finite subcover. And therefore $X$ is the finite union of adequate sets and it is adequate as well.

Finally, since rectangles form a basis for the product topology, from the lemma above we have that $X\times Y$ is indeed quasi-compact, as wanted. $\quad\square$

(I found the proof on ProofWiki sometime in the past.)

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    $\begingroup$ It looks like you are claiming that the $U$ you construct in the second paragraph is adequate, but I don't see it very clearly. In particular, there seems to be an implicit claim that $U \times X \subset W_1 \cup \cdots \cup W_n$. $\endgroup$ Apr 23, 2014 at 14:12
  • $\begingroup$ @Scott: Yes, I think you're right. What I should have here is that without loss of generality $W_i=U_i\times V_i$ (and we can make that assumption, even without choice). Then $U=\bigcap U_i$, and given some $(u,y)\in U\times Y$ there is some $V_i$ such that $y\in V_i$ and therefore $(u,y)\in U_i\times V_i$, so $U\times Y$ is adequate. $\endgroup$
    – Asaf Karagila
    Apr 23, 2014 at 14:24
  • $\begingroup$ I'll think about it for a few minutes to see if I can dispense of that assumption of rectangles, and I'll edit. (I also caught a use of "compact" instead of "quasi-compact" which I should correct while I'm at it.) $\endgroup$
    – Asaf Karagila
    Apr 23, 2014 at 14:24
  • $\begingroup$ @Scott: Yes, the use of rectangles is crucial here. Thanks, I'll edit it in. $\endgroup$
    – Asaf Karagila
    Apr 23, 2014 at 14:32

Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. We will prove that $Z$ is quasi-compact without using Axiom of Choice. Suppose $(W_\alpha)_{\alpha\in A}$ is an open cover of $Z$. Then $W_\alpha = \bigcup_{\beta\in B_\alpha} U_{\alpha\beta} \times V_{\alpha\beta}$, where $U_{\alpha\beta}$ is an open subset of $X$ and $V_{\alpha\beta}$ is an open subset of $Y$. Let $x \in X$. Then $\{x\}\times Y$ is quasi-compact. Since $(U_{\alpha\beta} \times V_{\alpha\beta})_{(\alpha,\beta)\in A\times B_\alpha}$ is an open cover of $\{x\}\times Y$, there exists a finite subcover $U_i(x) \times V_i(x), i = 1, \cdots , n_x$ of $\{x\}\times Y$, where each $U_i(x) \times V_i(x)$ is of the form $U_{\alpha\beta} \times V_{\alpha\beta}$ for some $(\alpha, \beta) \in A\times B_\alpha$. Let $U(x) = U_1(x) \cap \cdots \cap U_{n_x}$. Since $(U(x))_{x\in X}$ is an open cover of $X$, there exists a finite subcover $U(x_1),\cdots, U(x_m)$ of $X$. Then $U(x_i) \times V_k(x_i)$, $i = 1, \cdots, m, k = 1,\cdots, n_{x_i}$ is a finite cover of $X\times Y$. Then for every pair $(i, k)$, there exist $\alpha \in A$ and $\beta \in B_\alpha$ such that $U(x_i) \times V_k(x_i) \subset U_{\alpha\beta} \times V_{\alpha\beta} \subset W_\alpha$. Hence there exists a finite subcover of $(W_\alpha)_{\alpha\in A}$. QED

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    $\begingroup$ A problem with your write-up: For each $x$ there is a finite subcover of $\{x\}\times Y$, but there may be more than one, so you need to specify (if at all possible) how to pick the $U_i(x)\times V_i(x)$, $1\le i\le n_x$. $\endgroup$ Apr 14, 2014 at 16:32
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    $\begingroup$ @AndresCaicedo A good point. By the way, why did you vote to close this question? $\endgroup$ Apr 14, 2014 at 16:42

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